Question: Write the equation for a parabola with a focus at $(-6,0)$ and a directrix at $x=-2$. $x=$
Explanation: The strategy A parabola is defined as the set of all points that are the same distance away from a point (the focus) and a line (the directrix). Let $(x,y)$ be a point on the parabola. Then the distance between $(x,y)$ and the focus, $(-6,0)$, is equal to the distance between $(x,y)$ and the directrix, $x=-2$. Once we find these distances, we can equate them in order to derive the equation of our parabola. Finding the distances from $(x,y)$ to the focus and the directrix The distance between $(x,y)$ and $(-6,0)$ is $\sqrt{(x+6)^2+y^2}$. [How did we find that?] Similarly, the distance between $(x,y)$ and the line $x=-2$ is $\sqrt{(x+2)^2}$. [How did we know that?] Deriving the formula by equating the distances $\begin{aligned} \sqrt{(x+2)^2} &= \sqrt{(x+6)^2+y^2} \\\\ (x+2)^2 &= (x+6)^2+y^2 \\\\ {x^2}+4x{+4} &= {x^2}{+12x}+36+y^2\\\\ 4x{-12x}&=y^2+36{-4} \\\\ -8x&=y^2+32 \\\\ x&=-\dfrac{y^2}{8}-4\end{aligned}$ The answer The equation of our parabola is $x=-\dfrac{y^2}{8}-4$. Here is the graph of our parabola. As expected, the distance between a point on the parabola, $(x,y)$, and the focus is the same as the distance between $(x,y)$ and the directrix. ${1}$ ${2}$ ${3}$ ${4}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${\llap{-}10}$ ${\llap{-}11}$ ${\llap{-}12}$ ${\llap{-}13}$ ${\llap{-}14}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ $y$ $x$ ${(x,y)}$